Devlin's
Angle

July-August 2004

The Two Envelopes Paradox

I received a letter recently asking for
me to "rule" on a debate two people were
having about the notorious two envelopes
paradox. Since my efforts to convince
people of the correct resolution to the
Monty Hall Problem inevitably generate a
small avalanche of letters claiming I am
completely wrong, I have in the past
hesitated to tackle the much, much
trickier envelopes puzzle. But the time
has come, I think, to throw caution to
the wind, and enter the fray.

Here, for those unfamiliar with the
problem, is what it says.

You are taking part in a game show. The
host offers you two envelopes, each
containing a check. You may choose one,
keeping the money it contains. She tells
you that one envelope contains exactly
twice as much as the other, but does not
tell you which is which.

Since you have no way of knowing which
envelope contains the larger sum, you
pick one at random. The host asks you to
open the envelope. You do so and take out
a check for $40,000.

Here is where things get interesting,
especially for contestants who know some
mathematics.

The host now says you have a chance to
change your mind and choose the other
envelope. If you don't know anything
about probability theory, particularly
expectations, you probably say to
yourself, the odds are fifty-fifty that
you have chosen the larger sum, so you
may as well stick with your first choice.
(And you'd be right. But I'll come back
to this in a moment.)

On the other hand, if you know a bit
(though not too much) about probability
theory, you may well try to compute the
expected gain due to swapping. The
chances are you would argue as follows.
The other envelope contains either
$20,000 or $80,000, each with probability
.5. Hence the expected gain of swapping
is

[0.5 x 20,000] + [0.5 x 80,000] -
40,000 = 10,000

That's an expected gain of $10,000. So
you swap.

But wait a minute. There's nothing
special about the actual monetary amounts
here, provided one envelope contains
twice as much as the other. Suppose you
opened one envelope and found $M. Then
you would calculate your expected gain
from swapping to be

[0.5 x M/2] + [0.5 x 2M] - M = M/4

and since M/4 is greater than zero you
would swap. Right?

Okay, let's take this line of reasoning a
bit further. If it doesn't matter what M
is, then you don't actually need to open
the envelope at all. Whatever is in the
envelope you would choose to swap. Still
with me?

Well, if you don't open the envelope,
then you might as well choose the other
envelope in the first place. And having
swapped envelopes, you can repeat the
same calculation again and again,
swapping envelopes back and forward
ad-infinitum. There is no limit to the
cumulative expected gain you can obtain.
But this is absurd.

And there's the paradox. What is wrong
with the computation of the expected gain
from swapping?

The answer is everything. The above
computation is meaningless - which is why
it leads so easily to a nonsensical
outcome. If you want to apply probability
theory, you are free to do so, but you
need to do it correctly. And that means
working with actual probabilities, taking
care to distinguish between prior and
posterior probabilities. Let's take a
closer look.

As with the Monty Hall Problem, if you
really want to analyze the situation, you
have to start by looking at the way the
scenario was set up.

Let L denote the lower dollar value of
the two checks. The other check thus has
value 2L. Let P(L) be the prior
probability distribution for the choice
the host makes for the lower value in the
envelopes. (This will affect the entire
game. Of course, we don't know anything
about this distribution. But we can see
how it affects the outcome of the game.
Read on.

When you make your choice (C) during the
game, you choose either the envelope
containing the lower value (C=lower) or
the one that contains the higher
(C=higher). As the amounts are hidden
from you, you choose entirely at random,
with equal probabilities for the two
options, so

P(C=lower) = P(C=higher) = 0.5

During the game, the value (V) of the
content of the chosen envelope is
revealed to be a certain value M. Given
this information, what is the posterior
probability that the chosen envelope
contains the higher or lower value? That
is, what is P(C|V=M), the probability
that you chose the envelope containing
the lower/higher value, given you now
know what V is? This is the
probability you need in order to compute
any expected gain. The correct expected
gain calculation is:

(2M)P(C=lower|V=M)+(M/2)P(C=higher|V=M)
- M

The paradox above arose because you
assumed that

P(C=lower|V=M) = P(C=higher|V=M) = 0.5

Let's see why this cannot be the case.
(In what follows, remember that L, V, C
are variables and M is a numerical
constant.)

By Bayes' Theorem:

P(C|V=M) = P(V=M|C)P(C)/P(V=M) . . . (1)

Taking the first of the two cases, where
you choose the lower value (V=L), we have

P(V=M|C=lower) = P(L=M)

The second of the two cases, where the
chosen envelope contains double the lower
value, is

P(V=M|C=higher) = P(L=M/2)

Substituting each of these two identities
in to (1) gives

P(C=lower|V=M) = P(C=lower)P(L=M)/P(V=M)
. . . (2)

and

P(C=higher|V=M) =
P(C=higher)P(L=M/2)/P(V=M) . . . (3)

From (2), P(C=lower|V=M) is the same as
P(C=lower) only if P(L=M) is the same as
P(V=M). If it is not then in the
calculations of expected gain you have
used the incorrect probability.
Specifically, you have used the prior
probability, P(C=lower)=0.5, of choosing
the lower value, rather than the
posterior probability P(C=lower|V). The
same argument works for P(C=higher|V),
starting from (3).

Under what circumstances could we have
P(L=M) = P(V=M)?

Since P(V=M) must normalize the
distribution, we have

P(V=M) = P(C=lower)P(L=M) +
P(C=higher)P(L=M/2)

that is,

P(V=M) = 0.5 P(L=M) + 0.5 P(L=M/2) . .
. (4)

From (4), to have P(L=M) = P(V=M) we
would need P(L=M/2) = P(L=M) for all
values of M. This is an infinite uniform
"distribution". But no such distribution
exists (it cannot be normalised). Hence
it is impossible to have any prior
distribution for which P(L=M) = P(L=M/2)
is satisfied for all M. As a result there
is no posterior distribution for which,
given any V, we could have P(C=lower|V) =
0.5.

The result you will get in the game
depends on the prior distribution for the
amounts in the envelope. For example if
the prior distribution P(L) were uniform
between 0 and $30,000, and you found
$40,000 in the envelope, then you would
expect to lose if you swap, whereas if
the prior were uniform between $30,000
and $100,000, you would expect to gain.

To summarize: the paradox arises because
you use the prior probabilities to
calculate the expected gain rather than
the posterior probabilities. As we have
seen, it is not possible to choose a
prior distribution which results in a
posterior distribution for which the
original argument holds; there simply are
no circumstances in which it would be
valid to always use probabilities of 0.5.